# Palindrome Partitioning | DP-17

Given a string, a partitioning of the string is a palindrome partitioning if every substring of the partition is a palindrome. For example, “aba|b|bbabb|a|b|aba” is a palindrome partitioning of “ababbbabbababa”. Determine the fewest cuts needed for palindrome partitioning of a given string. For example, minimum 3 cuts are needed for “ababbbabbababa”. The three cuts are “a|babbbab|b|ababa”. If a string is palindrome, then minimum 0 cuts are needed. If a string of length n containing all different characters, then minimum n-1 cuts are needed. This problem is a variation of Matrix Chain Multiplication problem. If the string is palindrome, then we simply return 0. Else, like the Matrix Chain Multiplication problem, we try making cuts at all possible places, recursively calculate the cost for each cut and return the minimum value.

Let the given string be str and minPalPartion() be the function that returns the fewest cuts needed for palindrome partitioning. following is the optimal substructure property.

```// i is the starting index and j is the ending index. i must be passed as 0 and j as n-1
minPalPartion(str, i, j) = 0 if i == j. // When string is of length 1.
minPalPartion(str, i, j) = 0 if str[i..j] is palindrome.

// If none of the above conditions is true, then minPalPartion(str, i, j) can be
// calculated recursively using the following formula.
minPalPartion(str, i, j) = Min { minPalPartion(str, i, k) + 1 +
minPalPartion(str, k+1, j) }
where k varies from i to j-1
```

Following is Dynamic Programming solution. It stores the solutions to subproblems in two arrays P[][] and C[][], and reuses the calculated values.

## C/C++

 `// Dynamic Programming Solution for Palindrome Partitioning Problem ` `#include ` `#include ` `#include ` `  `  `// A utility function to get minimum of two integers ` `int` `min (``int` `a, ``int` `b) { ``return` `(a < b)? a : b; } ` `  `  `// Returns the minimum number of cuts needed to partition a string ` `// such that every part is a palindrome ` `int` `minPalPartion(``char` `*str) ` `{ ` `    ``// Get the length of the string ` `    ``int` `n = ``strlen``(str); ` `  `  `    ``/* Create two arrays to build the solution in bottom up manner ` `       ``C[i][j] = Minimum number of cuts needed for palindrome partitioning ` `                 ``of substring str[i..j] ` `       ``P[i][j] = true if substring str[i..j] is palindrome, else false ` `       ``Note that C[i][j] is 0 if P[i][j] is true */` `    ``int` `C[n][n]; ` `    ``bool` `P[n][n]; ` `  `  `    ``int` `i, j, k, L; ``// different looping variables ` `  `  `    ``// Every substring of length 1 is a palindrome ` `    ``for` `(i=0; i

## Java

 `// Java Code for Palindrome Partitioning  ` `// Problem ` `public` `class` `GFG  ` `{               ` `    ``// Returns the minimum number of cuts needed ` `    ``// to partition a string such that every  ` `    ``// part is a palindrome ` `    ``static` `int` `minPalPartion(String str) ` `    ``{ ` `        ``// Get the length of the string ` `        ``int` `n = str.length(); ` `       `  `        ``/* Create two arrays to build the solution ` `           ``in bottom up manner ` `           ``C[i][j] = Minimum number of cuts needed  ` `                     ``for palindrome partitioning ` `                     ``of substring str[i..j] ` `           ``P[i][j] = true if substring str[i..j] is ` `                     ``palindrome, else false ` `           ``Note that C[i][j] is 0 if P[i][j] is ` `           ``true */` `        ``int``[][] C = ``new` `int``[n][n]; ` `        ``boolean``[][] P = ``new` `boolean``[n][n]; ` `       `  `        ``int` `i, j, k, L; ``// different looping variables ` `       `  `        ``// Every substring of length 1 is a palindrome ` `        ``for` `(i = ``0``; i < n; i++) ` `        ``{ ` `            ``P[i][i] = ``true``; ` `            ``C[i][i] = ``0``; ` `        ``} ` `       `  `        ``/* L is substring length. Build the solution in ` `         ``bottom up manner by considering all substrings ` `         ``of length starting from 2 to n. The loop  ` `         ``structure is same as Matrx Chain Multiplication ` `         ``problem ( ` `        ``See https://tutorialspoint.dev/slugresolver/matrix-chain-multiplication-dp-8/ )*/` `        ``for` `(L = ``2``; L <= n; L++) ` `        ``{ ` `            ``// For substring of length L, set different ` `            ``// possible starting indexes ` `            ``for` `(i = ``0``; i < n - L + ``1``; i++) ` `            ``{ ` `                ``j = i + L - ``1``; ``// Set ending index ` `       `  `                ``// If L is 2, then we just need to  ` `                ``// compare two characters. Else need to ` `                ``// check two corner characters and value  ` `                ``// of P[i+1][j-1] ` `                ``if` `(L == ``2``) ` `                    ``P[i][j] = (str.charAt(i) ==  ` `                                ``str.charAt(j)); ` `                ``else` `                    ``P[i][j] = (str.charAt(i) ==  ` `                            ``str.charAt(j)) && P[i+``1``][j-``1``]; ` `       `  `                ``// IF str[i..j] is palindrome, then  ` `                ``// C[i][j] is 0 ` `                ``if` `(P[i][j] == ``true``) ` `                    ``C[i][j] = ``0``; ` `                ``else` `                ``{ ` `                    ``// Make a cut at every possible ` `                    ``// localtion starting from i to j, ` `                    ``// and get the minimum cost cut. ` `                    ``C[i][j] = Integer.MAX_VALUE; ` `                    ``for` `(k = i; k <= j - ``1``; k++) ` `                        ``C[i][j] = Integer.min(C[i][j],  ` `                                ``C[i][k] + C[k+``1``][j] + ``1``); ` `                ``} ` `            ``} ` `        ``} ` `       `  `        ``// Return the min cut value for complete  ` `        ``// string. i.e., str[0..n-1] ` `        ``return` `C[``0``][n-``1``]; ` `    ``} ` `       `  `    ``// Driver program to test above function ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `       ``String str = ``"ababbbabbababa"``; ` `       ``System.out.println(``"Min cuts needed for "``+ ` `                       ``"Palindrome Partitioning is "``+ ` `                          ``minPalPartion(str)); ` `    ``} ` `} ` `// This code is contributed by Sumit Ghosh `

## python3

 `# Dynamic Programming Solution for  ` `# Palindrome Partitioning Problem  ` ` `  `# Returns the minimum number of  ` `# cuts needed to partition a string ` `# such that every part is a palindrome ` `def` `minPalPartion(``str``): ` `     `  `    ``# Get the length of the string ` `    ``n ``=` `len``(``str``) ` `     `  `    ``# Create two arrays to build the  ` `    ``# solution in bottom up manner  ` `    ``# C[i][j] = Minimum number of cuts  ` `    ``#            needed for palindrome  ` `    ``#           partitioning of substring str[i..j]  ` `    ``# P[i][j] = true if substring str[i..j]  ` `    ``# is palindrome, else false. Note that ` `    ``# C[i][j] is 0 if P[i][j] is true  ` `    ``C ``=` `[[``0` `for` `i ``in` `range``(n)]  ` `            ``for` `i ``in` `range``(n)] ` `    ``P ``=` `[[``False` `for` `i ``in` `range``(n)]  ` `                ``for` `i ``in` `range``(n)] ` ` `  `    ``# different looping variables ` `    ``j ``=` `0` `    ``k ``=` `0` `    ``L ``=` `0` `     `  `    ``# Every substring of length  ` `    ``# 1 is a palindrome  ` `    ``for` `i ``in` `range``(n): ` `        ``P[i][i] ``=` `True``;  ` `        ``C[i][i] ``=` `0``;  ` `         `  `    ``# L is substring length. Build the  ` `    ``# solution in bottom up manner by  ` `    ``# considering all substrings of  ` `    ``# length starting from 2 to n.  ` `    ``# The loop structure is same as  ` `    ``# Matrix Chain Multiplication problem   ` `    ``# (See https://tutorialspoint.dev/slugresolver/matrix-chain-multiplication-dp-8/ ) ` `    ``for` `L ``in` `range``(``2``, n ``+` `1``): ` `         `  `        ``# For substring of length L, set  ` `        ``# different possible starting indexes  ` `        ``for` `i ``in` `range``(n ``-` `L ``+` `1``): ` `            ``j ``=` `i ``+` `L ``-` `1` `# Set ending index  ` `             `  `            ``# If L is 2, then we just need to ` `            ``# compare two characters. Else  ` `            ``# need to check two corner characters ` `            ``# and value of P[i+1][j-1] ` `            ``if` `L ``=``=` `2``:  ` `                ``P[i][j] ``=` `(``str``[i] ``=``=` `str``[j]) ` `            ``else``: ` `                ``P[i][j] ``=` `((``str``[i] ``=``=` `str``[j]) ``and`  `                             ``P[i ``+` `1``][j ``-` `1``]) ` `                              `  `            ``# IF str[i..j] is palindrome,  ` `            ``# then C[i][j] is 0 ` `            ``if` `P[i][j] ``=``=` `True``: ` `                ``C[i][j] ``=` `0` `            ``else``: ` `                 `  `                ``# Make a cut at every possible  ` `                ``# location starting from i to j, ` `                ``#and get the minimum cost cut. ` `                ``C[i][j] ``=` `100000000` `                ``for` `k ``in` `range``(i, j): ` `                    ``C[i][j] ``=` `min` `(C[i][j], C[i][k] ``+`  `                                   ``C[k ``+` `1``][j] ``+` `1``) ` `                                    `  `    ``# Return the min cut value for  ` `    ``# complete string. i.e., str[0..n-1]  ` `    ``return` `C[``0``][n ``-` `1``] ` ` `  `# Driver code ` `str` `=` `"ababbbabbababa"` `print` `(``'Min cuts needed for Palindrome Partitioning is'``,  ` `                                     ``minPalPartion(``str``)) ` `                                       `  `# This code is contributed  ` `# by sahil shelangia `

## C#

 `// C# Code for Palindrome Partitioning  ` `// Problem ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Returns the minimum number of cuts needed ` `    ``// to partition a string such that every  ` `    ``// part is a palindrome ` `    ``static` `int` `minPalPartion(String str) ` `    ``{ ` `        ``// Get the length of the string ` `        ``int` `n = str.Length; ` `         `  `        ``/* Create two arrays to build the solution ` `        ``in bottom up manner ` `        ``C[i][j] = Minimum number of cuts needed  ` `                    ``for palindrome partitioning ` `                    ``of substring str[i..j] ` `        ``P[i][j] = true if substring str[i..j] is ` `                    ``palindrome, else false ` `        ``Note that C[i][j] is 0 if P[i][j] is ` `        ``true */` `        ``int``[,] C = ``new` `int``[n, n]; ` `        ``bool``[,] P = ``new` `bool``[n, n]; ` `         `  `        ``int` `i, j, k, L; ``// different looping variables ` `         `  `        ``// Every substring of length 1 is a palindrome ` `        ``for` `(i = 0; i < n; i++) ` `        ``{ ` `            ``P[i, i] = ``true``; ` `            ``C[i, i] = 0; ` `        ``} ` `         `  `        ``/* L is substring length. Build the solution in ` `        ``bottom up manner by considering all substrings ` `        ``of length starting from 2 to n. The loop  ` `        ``structure is same as Matrx Chain Multiplication ` `        ``problem ( ` `        ``See https://tutorialspoint.dev/slugresolver/matrix-chain-multiplication-dp-8/ )*/` `        ``for` `(L = 2; L <= n; L++) ` `        ``{ ` `            ``// For substring of length L, set different ` `            ``// possible starting indexes ` `            ``for` `(i = 0; i < n - L + 1; i++) ` `            ``{ ` `                ``j = i + L - 1; ``// Set ending index ` `         `  `                ``// If L is 2, then we just need to  ` `                ``// compare two characters. Else need to ` `                ``// check two corner characters and value  ` `                ``// of P[i+1][j-1] ` `                ``if` `(L == 2) ` `                    ``P[i,j] = (str[i] == str[j]); ` `                ``else` `                    ``P[i,j] = (str[i] == str[j]) && ` `                             ``P[i + 1, j - 1]; ` `         `  `                ``// IF str[i..j] is palindrome, then  ` `                ``// C[i][j] is 0 ` `                ``if` `(P[i, j] == ``true``) ` `                    ``C[i, j] = 0; ` `                ``else` `                ``{ ` `                    ``// Make a cut at every possible ` `                    ``// localtion starting from i to j, ` `                    ``// and get the minimum cost cut. ` `                    ``C[i, j] = ``int``.MaxValue; ` `                    ``for` `(k = i; k <= j - 1; k++) ` `                        ``C[i, j] = Math.Min(C[i, j], C[i, k]  ` `                                  ``+ C[k + 1, j] + 1); ` `                ``} ` `            ``} ` `        ``} ` `         `  `        ``// Return the min cut value for complete  ` `        ``// string. i.e., str[0..n-1] ` `        ``return` `C[0, n - 1]; ` `    ``} ` `         `  `    ``// Driver program  ` `    ``public` `static` `void` `Main() ` `    ``{ ` `    ``String str = ``"ababbbabbababa"``; ` `    ``Console.Write(``"Min cuts needed for "``+ ` `                  ``"Palindrome Partitioning is "``+ ` `                  ``minPalPartion(str)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` `

Output:

`Min cuts needed for Palindrome Partitioning is 3 `

Time Complexity: O(n3)

An optimization to above approach
In above approach, we can calculate minimum cut while finding all palindromic substring. If we find all palindromic substring 1st and then we calculate minimum cut, time complexity will reduce to O(n2).
Thanks for Vivek for suggesting this optimization.

## C++

 `// Dynamic Programming Solution for Palindrome Partitioning Problem ` `#include ` `#include ` `#include ` `  `  `// A utility function to get minimum of two integers ` `int` `min (``int` `a, ``int` `b) { ``return` `(a < b)? a : b; } ` `  `  `// Returns the minimum number of cuts needed to partition a string ` `// such that every part is a palindrome ` `int` `minPalPartion(``char` `*str) ` `{ ` `    ``// Get the length of the string ` `    ``int` `n = ``strlen``(str); ` `  `  `    ``/* Create two arrays to build the solution in bottom up manner ` `       ``C[i] = Minimum number of cuts needed for palindrome partitioning ` `                 ``of substring str[0..i] ` `       ``P[i][j] = true if substring str[i..j] is palindrome, else false ` `       ``Note that C[i] is 0 if P[i] is true */` `    ``int` `C[n]; ` `    ``bool` `P[n][n]; ` `  `  `    ``int` `i, j, k, L; ``// different looping variables ` `  `  `    ``// Every substring of length 1 is a palindrome ` `    ``for` `(i=0; i

## Java

 `// Java Code for Palindrome Partitioning  ` `// Problem ` `public` `class` `GFG  ` `{             ` `    ``// Returns the minimum number of cuts needed ` `    ``// to partition a string such that every part ` `    ``// is a palindrome ` `    ``static` `int` `minPalPartion(String str) ` `    ``{ ` `        ``// Get the length of the string ` `        ``int` `n = str.length(); ` `       `  `        ``/* Create two arrays to build the solution ` `        ``in bottom up manner ` `           ``C[i] = Minimum number of cuts needed for ` `           ``palindrome partitioning of substring ` `           ``str[0..i] ` `           ``P[i][j] = true if substring str[i..j] is  ` `           ``palindrome, else false ` `           ``Note that C[i] is 0 if P[i] is true */` `        ``int``[] C = ``new` `int``[n]; ` `        ``boolean``[][] P = ``new` `boolean``[n][n]; ` `       `  `        ``int` `i, j, k, L; ``// different looping variables ` `       `  `        ``// Every substring of length 1 is a palindrome ` `        ``for` `(i = ``0``; i < n; i++) ` `        ``{ ` `            ``P[i][i] = ``true``; ` `        ``} ` `       `  `        ``/* L is substring length. Build the solution  ` `        ``in bottom up manner by considering all substrings  ` `        ``of length starting from 2 to n. */` `        ``for` `(L = ``2``; L <= n; L++) ` `        ``{ ` `            ``// For substring of length L, set different  ` `            ``// possible starting indexes ` `            ``for` `(i = ``0``; i < n - L + ``1``; i++) ` `            ``{ ` `                ``j = i + L - ``1``; ``// Set ending index ` `       `  `                ``// If L is 2, then we just need to  ` `                ``// compare two characters. Else need to  ` `                ``// check two corner characters and value ` `                ``// of P[i+1][j-1] ` `                ``if` `(L == ``2``) ` `                    ``P[i][j] = (str.charAt(i) == ` `                                 ``str.charAt(j)); ` `                ``else` `                    ``P[i][j] = (str.charAt(i) ==  ` `                           ``str.charAt(j)) && P[i+``1``][j-``1``]; ` `            ``} ` `        ``} ` `      `  `        ``for` `(i = ``0``; i < n; i++) ` `        ``{ ` `            ``if` `(P[``0``][i] == ``true``) ` `                ``C[i] = ``0``; ` `            ``else` `            ``{ ` `                ``C[i] = Integer.MAX_VALUE; ` `                ``for``(j = ``0``; j < i; j++) ` `                ``{ ` `                    ``if``(P[j+``1``][i] == ``true` `&& ``1` `+ ` `                                 ``C[j] < C[i]) ` `                        ``C[i] = ``1` `+ C[j]; ` `                ``} ` `            ``} ` `        ``} ` `       `  `        ``// Return the min cut value for complete ` `        ``// string. i.e., str[0..n-1] ` `        ``return` `C[n-``1``]; ` `    ``} ` `     `  `    ``// Driver program to test above function ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `       ``String str = ``"ababbbabbababa"``; ` `       ``System.out.println(``"Min cuts needed for "``+ ` `                          ``"Palindrome Partitioning"``+ ` `                          ``" is "``+ minPalPartion(str)); ` `    ``} ` `} ` `// This code is contributed by Sumit Ghosh `

## C#

 `// C# Code for Palindrome Partitioning  ` `// Problem ` `using` `System; ` ` `  `class` `GFG ` `{ ` `             `  `    ``// Returns the minimum number of cuts needed ` `    ``// to partition a string such that every part ` `    ``// is a palindrome ` `    ``static` `int` `minPalPartion(String str) ` `    ``{ ` `        ``// Get the length of the string ` `        ``int` `n = str.Length; ` `         `  `        ``/* Create two arrays to build the solution ` `        ``in bottom up manner ` `        ``C[i] = Minimum number of cuts needed for ` `        ``palindrome partitioning of substring ` `        ``str[0..i] ` `        ``P[i][j] = true if substring str[i..j] is  ` `        ``palindrome, else false ` `        ``Note that C[i] is 0 if P[i] is true */` `        ``int``[] C = ``new` `int``[n]; ` `        ``bool``[,] P = ``new` `bool``[n,n]; ` `         `  `        ``int` `i, j, L; ``// different looping variables ` `         `  `        ``// Every substring of length 1 is a palindrome ` `        ``for` `(i = 0; i < n; i++) ` `        ``{ ` `            ``P[i,i] = ``true``; ` `        ``} ` `         `  `        ``/* L is substring length. Build the solution  ` `        ``in bottom up manner by considering all substrings  ` `        ``of length starting from 2 to n. */` `        ``for` `(L = 2; L <= n; L++) ` `        ``{ ` `            ``// For substring of length L, set different  ` `            ``// possible starting indexes ` `            ``for` `(i = 0; i < n - L + 1; i++) ` `            ``{ ` `                ``j = i + L - 1; ``// Set ending index ` `         `  `                ``// If L is 2, then we just need to  ` `                ``// compare two characters. Else need to  ` `                ``// check two corner characters and value ` `                ``// of P[i+1][j-1] ` `                ``if` `(L == 2) ` `                    ``P[i,j] = (str[i] == str[j]); ` `                ``else` `                    ``P[i,j] = (str[i] == str[j]) && P[i+1,j-1]; ` `            ``} ` `        ``} ` `     `  `        ``for` `(i = 0; i < n; i++) ` `        ``{ ` `            ``if` `(P[0,i] == ``true``) ` `                ``C[i] = 0; ` `            ``else` `            ``{ ` `                ``C[i] = ``int``.MaxValue; ` `                ``for``(j = 0; j < i; j++) ` `                ``{ ` `                    ``if``(P[j+1,i] == ``true` `&& 1 + C[j] < C[i]) ` `                        ``C[i] = 1 + C[j]; ` `                ``} ` `            ``} ` `        ``} ` `         `  `        ``// Return the min cut value for complete ` `        ``// string. i.e., str[0..n-1] ` `        ``return` `C[n-1]; ` `    ``} ` `     `  `    ``// Driver program  ` `    ``public` `static` `void` `Main() ` `    ``{ ` `    ``String str = ``"ababbbabbababa"``; ` `    ``Console.Write(``"Min cuts needed for "``+ ` `                        ``"Palindrome Partitioning"``+ ` `                        ``" is "``+ minPalPartion(str)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` `

Output:

`Min cuts needed for Palindrome Partitioning is 3 `

Time Complexity: O(n2)